Determining the limit as the denominator goes to zero

Question

It's been a while since I've taken an intro calculus class. Could someone remind me of this? I'm guessing it's La'Hopitals but a refresher would be super helpful

$ \lim_{\gamma\to1} \frac{c_{t}^{1-\gamma}}{1-\gamma} = ln(C_t) $

your equation is wrong, the nominator goes to 0 the denominator to 1 so the expression goes to infty — trula, Oct 31 '19 at 02:29
hmmmm let me attach an image of what is written in the book I'm looking at in the book. Maybe I'm looking at it wrong? — financial_physician, Oct 31 '19 at 03:33

RRL · Accepted Answer · 2019-10-31T03:59:53.273

1

This is probably a misprint in the book. A power utility function should take the form

$$u(c_t) = \frac{c_t^{1-\gamma}-1}{1-\gamma}$$

Using L’Hopital’s rule for this $0/0$ indeterminate form, we have

$$\lim_{\gamma \to 1} u(c_t) = \lim_{\gamma \to 1}\frac{-\ln(c_t)e^{\ln(c_t)(1-\gamma)}}{-1} = \ln(c_t)$$

edited Oct 31 '19 at 03:59

answered Oct 31 '19 at 03:53

RRL

90,707

Also known as a CRRA or isoelastic utility function. See Wikipedia reference under isoelastic utility. – RRL Oct 31 '19 at 03:56
you're the man. Thank you – financial_physician Oct 31 '19 at 03:59
You’re very welcome. – RRL Oct 31 '19 at 04:03

Determining the limit as the denominator goes to zero

1 Answers1